Statistical Mechanics Homework 7

Transcription

1 Georgia Institute of Technology Statistical Mechanics Homework 7 Conner Herndon March 26, 206 Problem : Consider a classical system of N interacting monoatomic molecules at temerature T with Hamiltonian H ({r i }, { i }) K ({ i }) + U ({r i }), (.) where K is the kinetic energy and U is the interaction otential energy. (a) Show that the robability distribution for observing the system at hase sace oint ({r i }, { i }) factors into a osition-deendent art and a momentum-deendent art. Recall that the robability distribution for observing the system at hase sace oint ({r i }, { i }) is defined in terms of the hase sace density ρ ({r i }, { i }) as f ({r i }, { i }) ρ ({r i }, { i }) ρ ({ri }, { i }) d 3N rd 3N. (.2) (b) From here, show that the single-article momentum distribution function is { ( } 2 φ ( x, y, z ) (2πmkT ) ex x + 2 y + 2 z), (.3) where m is the molecule mass. This is usually called the Maxwell-Boltzmann distribution. It is the correct momentum distribution function for a article of mass m in a thermally equilibrated system. The system can be in any hase (gas, liquid, or crystal), and the distribution is still valid rovided classical mechanics is accurate. One consequence is that the average seed (or momentum) of a article is the same in a liquid and a gas, rovided the temerature is the same. Of course, the frequency of collisions in a liquid is much higher than in a gas. For this reason, a molecule will travel much farther er unit time in a gas hase than in a condensed hase even though the single-molecule velocity distributions are identical in the two hases.

2 (c) Show that the fraction of molecules with momentum between and + d is 4π 2 } φ () d { (2πmkT ) ex 2 d. (.4) (d) Prove that the most robable molecular seed is v 2kT/m. (.5) (e) Prove that the mean seed is Note that v v. v 8kT πm. (.6) (f) Show that the mean square fluctuation of the seed is v 2 v 2 kt ( 3 8 ). (.7) m π Note that ( v) 2 α v 2, with α 0.8. Hence, the mean square fluctuation of the seed is comarable to the mean square seed. (g) Show that the artition function for the N-article system factors in such a way that Q (N, V, T ) Q ideal Q con, where Q ideal is the ideal gas artition function and Q con V N d 3N re βu({r i}) (.8) is the so-called configurational artition function. The hase sace density is roortional to the Boltzmann weight ρ ({r i }, { i }) e βh ({r i},{ i }). (.9) Then the robability distribution for observing the system at hase sace oint ({r i }, { i }) is e βh ({r i},{ i }) f ({r i }, { i }) e βh ({r i },{ i }) d 3N rd 3N ( ) ( ) e βk({ i }) e βu{ri} d 3N rd 3N e βk({i}) e βu{r i} ( ) ( ) e βk({ i}) e βu{r i} e βk({ i }) d 3N e βu{r i } d 3N r (.0) We have then broken the robability density into momentum- and osition-deendent arts.

3 (b) Letting K 2 /2m, β /kt, and 2 2 x + 2 y + 2 z, The momentum-deendent art is φ ( x, y, z ) e ( 2 x +2 y +2 z). (.) e (2 x +2 y +2 z) d x d y d z The denominator integral may be slit into the roduct of three identical integrals with the value e s2 /α ds απ, (.2) giving the single-article momentum distribution function { ( } 2 φ ( x, y, z ) (2πmkT ) ex x + 2 y + 2 z). (.3) (c) To find the fraction of articles ossessing between momentum and + d, we consider a sace sanned by x, y, and z. The number of articles with momentum between and + d is equal to the number of articles contained within the ball of radius. Then the fraction of articles is equal to the single-article momentum distribution function multilied by the solid angle and radial differential element ( { ( φ () d 4π 2 2 ) }) (2πmkT ) ex d 4π 2 (2πmkT ) ex { 2 } d. (.4) (d) The most robably velocity v may be found by taking the maximum of φ (), 0 φ () 8πe 2 /( ) + ( 4π 2) ( 2 2, ) e 2 /( ) (.5) giving and therefore v 2kT m m. (.6) (e) On the other hand, the mean seed is

4 v m ( ) φ () d m φ () d ( ) 3 e 2 / d m 2 e 2 / d (.7) 8 m π mkt 8kT πm. (f) To get the mean square fluctuation, we also need to mean squared momentum v 2 m 2 2 ( ) 4 e 2 / d m 2 2 e 2 / d m 2 (3mkT ) 3kT m. (.8) Then the mean square fluctuation of seed is v 2 v 2 3kT m 8kT πm kt ( 3 8 ). m π (.9) (g) The artition function is Q (N, V, T ) N!h 3N N!h 3N N e βh d 3 r i d 3 i i e βu({ri}) d 3N r e i 2 i (2πkT )3N/2 e βu({ri}) d 3N r N!h3N Q ideal e βu({ri}) d 3N r V N Q ideal Q con. N d 3 i i (.20)

5 Problem 2: Consider a system of N non-interacting dioles,, in an external magnetic field B Bẑ. (a) Obtain the artition function using classical statistical mechanics. Note that in this case, all - orientations are allowed. (b) Obtain al z. Comare the result with that obtained using the quantum mechanical treatment and discuss what you learn from the comarison. The energy of a diole is E B B cos θ. (2.) Then the single-diole artition function is Q (V, T ) 2π π 0 0 e βb cos θ sin θdθdφ 4π sinh (βb). βb (2.2) Since the dioles are noninteracting, the N diole artition function is the single-diole artition function raised to the ower N Q (N, V, T ) N! The average z-comonent of magnetic moment may be found by ( ) 4π N sinh N (βb). (2.3) βb z N z. (2.4) By the Gibbs equation de T ds + z db, (2.5) and by Legendre transform the Helmholtz is da SdT z db, (2.6) meaning z ( ) A. (2.7) B T By the micro-macro connection A kt log Q, so

6 ( [ 4π z kt B log βb kt B sinh (βb) B kt B sinh (βb) ]) sinh (βb) [ sinh (βb) B [ β sinh (βb) + cos (βb) B2 B ] T ] (2.8) Then the al mean magnetic moment is coth (βb) kt B. In the quantum mechanical treatment, we had {( z QM Ng BJ + ) coth 2J NkT z N coth (βb) B. (2.9) [( + 2J ) ] βg B JB [ ]} 2J coth 2 βg BB. (2.0) If we hold constant while letting J (meaning g 0), then the two functions are the same this fact reflects the classical limit of quantum mechanics when the sin degrees of freedom reach a continuum. In the limit where the field is much greater than the temerature, z classical N z QM Ng BJ. (2.) On the other hand, in the limit when the temerature is much greater than the external field, z classical N2 B 3kT z QM Ng2 2 B J (J + ) B 3kT N2 B 3kT. (2.2) This high temerature limit reflects Curie s law, and the quantum mechanical treatment secifies the form of Curie s constant when taking the artial ( ) z χ T lim C B 0 B T. (2.3) T